 Solution

Write $b=(\ell +1)a$, $R(r)=r^{\ell }{e}^{-r/b}$ and $Y(\theta ,\varphi )={\sin }^{\ell }\theta e^{i\ell \varphi }$, for nonnegative integer $\ell$. The angular Laplacian satisfies ${\Delta }_{S^2}Y=-\ell (\ell +1)Y$. Direct radial differentiation givesBecause $(\ell +1)/b=1/a$, the $1/r$ contribution cancels the Coulomb potential in the Hamiltonian. ThereforeThe volume element contributes $r^2\sin \theta$. The squared state factors into radial, polar and uniform azimuthal densities, so the angular normalization cancels in every radial expectation. With $I_j={\int }_0^{\infty }{r}^j{e}^{-2r/b}dr$,The supplied integration-by-parts recurrence gives $I_{j+1}=(j+1)b{I}_j/2$. HenceIn particular $\mathrm{Var}(r)/{\mu }^2=(2\ell +3{)}^{-1}$. Put $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality gives

For the polar variable set $J_j={\int }_0^{\pi }{\sin }^j\theta d\theta$. The polar density is proportional to ${\sin }^{2\ell +1}\theta$, soAn exact expression for the requested tail, as well as its useful estimate, isThe inequality is Markov's inequality applied to ${\cos }^2\theta$. If a finite expression is desired, expand $(1-z^2{)}^{\ell }$: the numerator is ${\sum }_{j=0}^{\ell }(-1{)}^j(\genfrac{}{}{0ex}{}{\ell }{j})(1-h^{2j+1})/(2j+1)$ and the denominator is the same sum with $h=0$.

By the union bound, with probability at least $1-2h$ both $|r/\mu -1|\le h$ and $|\cos \theta |\le h$. The height above the equatorial plane is then at most $(1+h)\mu h$. The cylindrical radius differs from $\mu$ by at most $\mu h+(1+h)\mu h^2$. Thus the distance from the circle of radius $\mu$ in the equatorial plane is $O(\mu h)$ with probability tending to one. This is relative concentration around a growing circle; it does not assert an absolutely shrinking thickness. The probability concentration of a circular Coulomb eigenstate is a precise version of the large-angular-momentum orbit picture.